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Is the graph of $f(x)=|x|\,x$ (or any $C^1$ function that is not $C^\infty)$ a smooth embedded submanifold of $\mathbb{R}^2$ with its standard differential structure?

I apologize if this is too elementary, but I got somehow confused. I understand that the question is equivalent to the existence of a diffeomorphism of a neighborhood $U$ of $(0,0)\in\mathbb{R}^2$ and $\mathbb{R}^2$ that takes the graph of $f$ to a straight line.

Daniel Fischer
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Peter Franek
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    And by smooth you mean $C^{\infty}$. No, and it's a good exercise. – orangeskid Jan 24 '15 at 11:11
  • The question was implicitely answered negatively as a reply to a similar question of mine here, so I vote to close this. – Peter Franek Jan 24 '15 at 11:14
  • Here is a hint: a $C^{k}$ submanifold is a graph of a $C^{k}$ function in almost all linear coordinate systems. You get $\alpha x + f(x)$ $C^{\infty}$ for almost all $\alpha$ ... – orangeskid Jan 24 '15 at 11:20
  • It's your own question, and there are no answers. If you want to delete it, you can do so on your own (that is, you don't need mods to do it for you) – davidlowryduda Jan 24 '15 at 19:31

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